Answer
$$0$$
Work Step by Step
Given $$\lim_{x\to 0} \frac{\sin^2kx }{x},\ \ k\neq0 $$
Then \begin{align*}
\lim_{x\to 0} \frac{\sin^2kx }{x}&=(\lim_{x\to 0} \sin kx)\left( \lim_{x\to 0} \frac{k\sin kx }{kx}\right)\\
&= (0)(k)\\
&=0
\end{align*}
Calculus, 10th Edition (Anton)
by Anton, Howard
Published by
Wiley
ISBN 10:
0-47064-772-8
ISBN 13:
978-0-47064-772-1
Chapter 1 - Limits and Continuity - 1.6 Continuity of Trigonometric Functions - Exercises Set 1.6 - Page 106: 32
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